Group Cohomology as a Derived Functor — Examples

Degree zero cochains are $M$, while every positive normalized bar is zero because its only group entry is $1$.

The normalized complex is $M$ in degree zero and zero above it, proving the computation.

Fixed points satisfy $m=-m$, hence $m=0$ in $\mathbb{Z}$.

Coinvariants quotient by $tm-m=-2m$, hence are $\mathbb{Z}\mathrm{/}2$; degree-zero recovery identifies these with the two asserted groups.

Substitution in the definition yields the two displayed formulas and $d(g_0)=0$.

Applying $d$ to the three terms of the second formula produces each vertex twice with opposite signs, explicitly illustrating $d^2=0$.

Set $g=h=1$ to get $z(1)=z(1)+z(1)$.

Subtracting $z(1)$ gives $z(1)=0$. Thus the normalized representative is $z$ itself.

$(t-1)N=N(t-1)=t^m-1=0$, so this is a complex.

Writing an element as $\sum a_i{t}^i$ shows $\ker \epsilon =(t-1)\mathbb{Z}[C_m]$, $\ker (t-1)=N\mathbb{Z}[C_m]$, and $\ker N=(t-1)\mathbb{Z}[C_m]$. Hence it is exact and every term is free.

Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_m]}(-,M)$ identifies every cochain group with $M$; the coboundaries alternate between $t-1$ and $N$.

Taking kernel modulo preceding image gives exactly the displayed groups, including the degree-zero kernel.

Induction and coinduction from $1$ are respectively $\mathbb{Z}[G]\otimes M$ and ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}[G],M)$.

Shapiro identifies their positive (co)homology with positive (co)homology of the trivial group, which vanishes.

The contraction has $s(1)=(1,1)$, so $hs(1)=(h,h)$.

But $s(h\cdot (1))=s(h)=(1,h)$, which differs from $(h,h)$. Thus equivariance fails.

The trivial module for $1$ is free, so its projective dimension is zero. For $\mathbb{Z}$, $0\to \mathbb{Z}[t,t^{-1}]\stackrel{t-1}{\to }\mathbb{Z}[t,t^{-1}]\to \mathbb{Z}\to 0$ is a free resolution.

With trivial coefficients $\mathbb{Z}$, applying Hom sends $t-1$ to zero, so $H^1(\mathbb{Z};\mathbb{Z})=\mathbb{Z}\mathrm{\ne }0$. Hence the resolution length one is minimal.